Question

The value of
\(\int^1_0\int_0^{1-x}\cos(x^3+y^2)dy\ dx-\int^1_0\int_0^{1-x}\cos(x^3+y^2)dx\ dy\)
is

• A. 0
• B. \(\frac{\cos(1)}{2}\)
• C. \(\frac{\sin(1)}{2}\)
• D. \(\cos(\frac{1}{2})-\sin(\frac{1}{2})\)

Answer 1

The Correct Option is A

To solve the given problem, we need to evaluate the expression: 

\(\int^1_0\int_0^{1-x}\cos(x^3+y^2)dy\ dx-\int^1_0\int_0^{1-x}\cos(x^3+y^2)dx\ dy\)

Let's break down the given expression into two double integrals and understand the regions of integration:

1. The first integral is:
2. The second integral is:
3. Switching the order of integration of the first integral, while considering the region:
4. Reflecting this region through \(y=x\):
   • For fixed \(y\), \(x\) varies from 0 to \(1-y\).
   • For fixed \(x\), \(y\) varies from 0 to \(1-x\).

By Fubini's theorem for changing orders of integration, the two integrals are equal:

\(\int_0^1 \int_0^{1-x} \cos(x^3 + y^2) \, dy\, dx = \int_0^1 \int_0^{1-y} \cos(x^3 + y^2) \, dx\, dy\)

Thus, the difference of these two integrals is:

\(\int^1_0\int_0^{1-x}\cos(x^3+y^2)dy\ dx - \int^1_0\int_0^{1-x}\cos(x^3+y^2)dx\ dy = 0\)

Conclusion: The given expression evaluates to 0, hence the correct answer is 0.